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3 years ago

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A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.6 cm before reaching its new equilibrium

length. The block is then pulled down slightly and released. What is the frequency of oscillation?

1 answer:

Kaylis [27]3 years ago

7 0

Answer:

f = 3.09 Hz

Explanation:

This is a simple harmonic motion exercise where the angular velocity is

w² = $\frac{k}{m}$

to find the constant (k) of the spring, we use Hooke's law with the initial data

F = - kx

where the force is the weight of the body that is hanging

F = W = m g

we substitute

m g = - k x

k = $- \frac{m g}{x}$

we calculate

k = $- \frac{9.8 m}{- 2.6 \ 10^{-2}}$

k = 3.769 10² m

we substitute in the first equation

w² = $\frac{ 3.769 \ 10^2 \ m }{m}$

w = 19.415 rad / s

angular velocity and frequency are related

w = 2πf

f = $\frac{w}{2\pi }$

f = 19.415 / 2pi

f = 3.09 Hz

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Let us now tackle the problem!

<u>Given:</u>

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<u>Unknown:</u>

Velocity of Rotation = v = ?

<u>Solution:</u>

$F = ma$

$F = m\frac{v^2}{R}$

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<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437

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Acceleration : brainly.com/question/2283922

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Trajectory Motion : brainly.com/question/8656387

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Circular , Ball , Centripetal

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